1. Infinite Limit Proof

I need to prove:

lim (x^4)-(2(x^2)) = +inf
x->+inf

Any suggestions?

2. Hi,

Hint : $\displaystyle x^4-2x^2=x^4\left(1-\frac{2}{x^2}\right)$

3. I don't know how to do it. The problem is done and over with, but I still would like to know how to get the answer for my own knowledge.

4. Originally Posted by Calc1stu
I don't know how to do it. The problem is done and over with, but I still would like to know how to get the answer for my own knowledge.
$\displaystyle \lim_{x\to\infty}x^4\left(1-\frac{2}{x^2}\right)=\lim_{x\to\infty}x^4\cdot\lim _{x\to\infty}\left(1-\frac{2}{x^2}\right)=\infty\cdot{1}=\infty$

5. It was supposed to be an epsilon delta proof.

6. You want to show that for every $\displaystyle M > 0$, there exists $\displaystyle N > 0$ such that whenever $\displaystyle x > N$, it necessarily follows that $\displaystyle f(x) > M \Leftrightarrow x^4 - 2x^2 > M$

Consider the latter expression as an equality:
$\displaystyle x^4 - 2x^2 = M$

$\displaystyle \Leftrightarrow x^4 - 2x^2 - M = 0$

$\displaystyle \Leftrightarrow x^2 = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-M)}}{2}$

$\displaystyle \Leftrightarrow x^2 = \frac{2 \pm \sqrt{4 + 4M}}{2}$

$\displaystyle \Leftrightarrow x^2 = 1 \pm \sqrt{1 + M}$

$\displaystyle \Leftrightarrow x^2 = 1 + \sqrt{1 + M}$ (since $\displaystyle x^2 \geq 0$ and $\displaystyle M > 0$)

$\displaystyle \Leftrightarrow x = +\sqrt{1 + \sqrt{1 + M}}$ (we're considering positive x's)

So when $\displaystyle x \ {\color{red}=} \ \sqrt{1 + \sqrt{1 + M}}$ we have that $\displaystyle f(x)\ {\color{red} =}\ M$. We want to show that for some $\displaystyle x \ {\color{magenta}>}\ N$, we have that $\displaystyle f(x) \ {\color{magenta}>}\ M$

Now consider: $\displaystyle x \ {\color{magenta}> }\ \sqrt{1 + \sqrt{1 + M}} > 0$. Can you finish?

7. As far as epsilon and delta are concerned which is M and N. I don't know where to go from there. All we ever usually do or did was a basic function.